 Alan Smith

2021-12-27

How do you simplify $\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$? Paul Mitchell

Explanation:
Knowing that
${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)=1$
We divide both sides by ${\mathrm{sin}}^{2}\left(\theta \right)$ so we have
$1+{\mathrm{cot}}^{2}\left(\theta \right)={\mathrm{csc}}^{2}\left(\theta \right)$
Or,
$1+\frac{1}{{\mathrm{tan}}^{2}\left(\theta \right)}=\frac{1}{{\mathrm{sin}}^{2}\left(\theta \right)}$
Taking the least common multiple we have
$\frac{{\mathrm{tan}}^{2}\left(\theta \right)+1}{{\mathrm{tan}}^{2}\left(\theta \right)}=\frac{1}{{\mathrm{sin}}^{2}\left(\theta \right\}}$
Inverting both sides we have
${\mathrm{sin}}^{2}\left(\theta \right)=\frac{{\mathrm{tan}}^{2}\left(\theta \right)}{{\mathrm{tan}}^{2}\left(\theta \right)+1}$
So we say that $\theta =\mathrm{arctan}\left(x\right)$
${\mathrm{sin}}^{2}\left(\mathrm{arctan}\left(x\right)\right)=\frac{{\mathrm{tan}}^{2}\left(\mathrm{arctan}\left(x\right)\right)}{{\mathrm{tan}}^{2}\left(\mathrm{arctan}\left(x\right)\right)+1}$
Knowing that $\mathrm{tan}\left(\mathrm{arctan}\left(x\right)\right)=x$
${\mathrm{sin}}^{2}\left(\mathrm{arctan}\left(x\right)\right)=\frac{{x}^{2}}{{x}^{2}+1}$
So we take the square root of both sides
$\mathrm{sin}\left(\mathrm{arctan}\left(x\right)\right)=±\sqrt{\frac{{x}^{2}}{{x}^{2}+1}}=±\frac{|x|}{\sqrt{{x}^{2}+1}}$ Deufemiak7

We can use the principles of "SOH-CAH-TOA".
First, let's call $\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(x\right)\right)=\mathrm{sin}\left(\theta \right)$ where the angle $\theta ={\mathrm{tan}}^{-1}\left(x\right)$
More specifically, ${\mathrm{tan}}^{-1}\left(x\right)=\theta$ is the angle when $\mathrm{tan}\left(\theta \right)=x$. We know this from the definition of inverse functions.
Since $\mathrm{tan}\left(\theta \right)=\frac{opposite}{adjacent}$, and here $\mathrm{tan}\left(\theta \right)=\frac{x}{1}$ we know that
$\left\{\begin{array}{c}Opposite=x\\ adjacent=1\\ hypotenuse=?\end{array}$
Using the Pythagorean Theorem, we can see that the hypotenuse of a right triangle with legs x and 1 has hypotenuse $=\sqrt{{x}^{2}+1}$
Now find $\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(x\right)\right)$
Since $\mathrm{sin}\left(\theta \right)=\frac{opposite}{hypotenuse}$, we see that
$\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(x\right)\right)=\frac{x}{\sqrt{{x}^{2}+1}}$ karton